Exercise 2.7

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Deduce from Ex. 2.6 that $\ord_p n! \leq n/(p - 1)$ and that $\sqrt[n]{n!} \leq \prod_{p \leq n} p^{1/(p-1)}$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

(The original statement $\prod_{p \mid n} p^{1/(p-1)}$ was modified.)
\begin{proof}
\begin{align*}
\ord_p\, n!  &= \sum_{k\geq 1} \left \lfloor \frac{n}{p^{k}}ight floor \leq  \sum_{k\geq 1}  \frac{n}{p^{k}} = \frac{n}{p} \frac{1}{1 - \frac{1}{p}} = \frac{n}{p-1}
\end{align*}

The decomposition of $n!$ in prime factors is

$n! = p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$ 
where $\alpha_i = \ord_{p_i}\, n! \leq \frac{n}{p_i-1}$, and $p_i \leq n, \ i=1,2,\cdots,k$.

Then
\begin{align*}
n! &\leq p_1^{\frac{n}{p_1-1}}p_2^{\frac{n}{p_2-1}}\cdots p_k^{\frac{n}{p_n-1}}\
\sqrt[n]{n!} &\leq p_1^{\frac{1}{p_1-1}}p_2^{\frac{1}{p_2-1}}\cdots p_k^{\frac{1}{p_n-1}}\
&\leq \prod_{p\leq n} p^{\frac{1}{p-1}}
\end{align*}
(the values of $p$ in this product describe all prime numbers $p\leq n$.)
\end{proof}

Exercise 2.7

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Exercise 2.7

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